Average Acceleration and Instantaneous Acceleration — Question 2

Question 2

Differentiate a spatial position history twice

Let $u=t/(1\ \mathrm s)$ and $\vec r=(u^2,2u^3,3u)$ m. Define $\vec v=d\vec r/dt$ and $\vec a=d\vec v/dt$.

1. Find v and a at 1 s.

2. Find average acceleration from 0 to 2 s and compare it with the instantaneous value at the midpoint.

3. Explain why the z component of acceleration is zero despite continuing vertical motion.

Question 2 - Solution

Differentiate a spatial position history twice

(a) Differentiate each component with its time scale.

$$\vec v=(2u,6u^2,3)\,\mathrm{m/s},\quad\vec a=(2,12u,0)\,\mathrm{m/s^2},\quad\boxed{\vec a(1)=(2,12,0)\,\mathrm{m/s^2}}$$ $\vec v=(2u,6u^2,3)$ m/s and $\vec a=(2,12u,0)$ m/s$^2$. At 1 s they are $(2,6,3)$ m/s and $(2,12,0)$ m/s$^2$. Acceleration has one additional inverse-time factor relative to velocity.

(b) Use the two endpoint velocities.

$$\vec a_{\mathrm{av}}=\frac{(4,24,3)-(0,0,3)}2=(2,12,0)\,\mathrm{m/s^2}=\vec a(1)$$ $\vec v(0)=(0,0,3)$ and $\vec v(2\ \mathrm s)=(4,24,3)$ m/s. Dividing their difference by 2 s gives $(2,12,0)$ m/s$^2$, equal to $\vec a(1\ \mathrm s)$. Here acceleration is linear in time, so its interval mean equals its midpoint value.

(c) Zero acceleration means unchanged velocity component.

$$z=3u\,\mathrm m,\quad v_z=3\,\mathrm{m/s},\quad a_z=\frac{dv_z}{dt}=0$$ $v_z=3$ m/s is constant, not zero. Hence z increases steadily while $a_z=0$. A zero component of acceleration constrains how the corresponding velocity changes; it does not say that the coordinate is fixed.